Problem 45
Question
In each pair of bonds, predict which is shorter. (a) \(\mathrm{B}-\mathrm{Cl}\) or \(\mathrm{Ga}-\mathrm{Cl}\) (b) \(\mathrm{Sn}-\mathrm{O}\) or \(\mathrm{C}-\mathrm{O}\) (c) \(\mathrm{P}-\mathrm{S}\) or \(\mathrm{P}-\mathrm{O}\) (d) \(\mathrm{C}=\mathrm{O}\) or \(\mathrm{C}=\mathrm{N}\)
Step-by-Step Solution
Verified Answer
(a) B-Cl is shorter, (b) C-O is shorter, (c) P-O is shorter, (d) C=O is shorter.
1Step 1: Analyze Atomic Size
Compare the atomic sizes of the central atoms in each pair of bonds because smaller atoms generally form shorter bonds. For (a), B is smaller than Ga. For (b), C is smaller than Sn. For (c), O is smaller than S.
2Step 2: Consider Bond Order
Higher bond orders typically lead to shorter bonds due to increased electron sharing. For (d), C=O and C=N both have a double bond, so this factor may not significantly differentiate them directly.
3Step 3: Compare Bond Length Trends
For (a), the smaller size of B compared to Ga suggests that B-Cl will be shorter than Ga-Cl. For (b), C-O, involving smaller C compared to Sn, will be shorter than Sn-O. For (c), due to O being smaller and having stronger bonds generally, P-O is likely shorter than P-S. In (d), due to oxygen's higher electronegativity, C=O is typically shorter than C=N.
Key Concepts
Atomic SizeBond OrderElectronegativityChemical Bonding
Atomic Size
Atomic size refers to the dimensions of an atom, usually measured by the atomic radius. It plays a crucial role in determining bond length, which is the distance between the centers of two bonded atoms. Atoms with smaller atomic sizes tend to form shorter bonds because their nuclei are closer together, allowing the shared electrons to exert a stronger binding force.
For instance:
For instance:
- Boron (B) is smaller than Gallium (Ga), which means a boron-chlorine (B-Cl) bond is shorter than a gallium-chlorine (Ga-Cl) bond.
- Carbon (C) has a smaller atomic size compared to Tin (Sn), hence a carbon-oxygen (C-O) bond is shorter than a tin-oxygen (Sn-O) bond.
- Similarly, Oxygen (O) is smaller than Sulfur (S), so phosphorus-oxygen (P-O) bonds are shorter than phosphorus-sulfur (P-S) bonds.
Bond Order
Bond Order describes the number of chemical bonds between a pair of atoms. It provides insight into the stability and length of a bond. Higher bond orders indicate more shared electron pairs, thus stronger and typically shorter bonds.
Consider this:
Consider this:
- Single bonds (bond order 1) are generally longer than double bonds (bond order 2) and triple bonds (bond order 3). This is due to the fact that more shared electrons bring the atoms closer together.
- In the example of C=O and C=N from the exercise, both bonds are double bonds, implying that only bond order wouldn't distinguish their lengths significantly unless considering other factors like electronegativity.
Electronegativity
Electronegativity refers to an atom's ability to attract and hold electrons within a chemical bond. It can affect bond length indirectly, especially when comparing bonds of the same order, such as those between carbon-oxygen (C=O) and carbon-nitrogen (C=N).
Points to note:
Points to note:
- Oxygen is more electronegative than nitrogen. This means it attracts electrons more strongly, creating a tighter bond. As a result, a C=O bond is usually shorter than a C=N bond.
- Higher electronegativity can lead to shorter bond lengths because the highly electronegative atom pulls shared electrons closer.
Chemical Bonding
Chemical bonding refers to the lasting attraction between atoms that enables the formation of chemical compounds. Bonds can be ionic, covalent, or metallic, but in the context of bond length, we focus on covalent bonds, where electrons are shared between atoms.
Covalent bond length depends on:
Covalent bond length depends on:
- Atomic Size: Smaller atoms form shorter covalent bonds due to closer nuclei.
- Bond Order: More bonds (higher bond order) mean shorter distances between bonded atoms as they share more electrons.
- Electronegativity: Bonds with highly electronegative atoms can have shorter distances due to tighter electron attractions.
Other exercises in this chapter
Problem 43
Give the bond order for each bond in the following molecules or ions: (a) \(\mathrm{CH}_{2} \mathrm{O}\) (b) \(\mathrm{SO}_{3}^{2-}\) (c) \(\mathrm{NO}_{2}^{+}\
View solution Problem 44
Give the bond order for each bond in the following molecules or ions: (a) \(\mathrm{CN}^{-}\) (b) \(\mathrm{CH}_{3} \mathrm{CN}\) (c) \(\mathrm{SO}_{3}\) (d) \(
View solution Problem 46
In each pair of bonds, predict which is shorter. (a) \(\mathrm{Si}-\mathrm{N}\) or \(\mathrm{Si}-\mathrm{O}\) (b) \(\mathrm{Si}-\mathrm{O}\) or \(\mathrm{C}-\ma
View solution Problem 47
Consider the nitrogen-oxygen bond lengths in \(\mathrm{NO}_{2}^{+}, \mathrm{NO}_{2}^{-},\) and \(\mathrm{NO}_{3}^{-} .\) In which ion is the bond predicted to b
View solution